Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{k + 6}{k^2 - 13k + 40} \div \dfrac{-3k^2 + 3k}{2k^3 - 12k^2 + 10k} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{k + 6}{k^2 - 13k + 40} \times \dfrac{2k^3 - 12k^2 + 10k}{-3k^2 + 3k} $ First factor out any common factors. $n = \dfrac{k + 6}{k^2 - 13k + 40} \times \dfrac{2k(k^2 - 6k + 5)}{-3k(k - 1)} $ Then factor the quadratic expressions. $n = \dfrac {k + 6} {(k - 5)(k - 8)} \times \dfrac {2k(k - 5)(k - 1)} {-3k(k - 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {(k + 6) \times 2k(k - 5)(k - 1) } { (k - 5)(k - 8) \times -3k(k - 1)} $ $n = \dfrac {2k(k - 5)(k - 1)(k + 6)} {-3k(k - 5)(k - 8)(k - 1)} $ Notice that $(k - 5)$ and $(k - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {2k\cancel{(k - 5)}(k - 1)(k + 6)} {-3k\cancel{(k - 5)}(k - 8)(k - 1)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $n = \dfrac {2k\cancel{(k - 5)}\cancel{(k - 1)}(k + 6)} {-3k\cancel{(k - 5)}(k - 8)\cancel{(k - 1)}} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $n = \dfrac {2k(k + 6)} {-3k(k - 8)} $ $ n = \dfrac{-2(k + 6)}{3(k - 8)}; k \neq 5; k \neq 1 $